# If

Question:

If $x^{3} d y+x y d x=x^{2} d y+2 y d x ; y(2)=e$ and $x>1$, then $y(4)$ is equal to :

1. (1) $\frac{3}{2}+\sqrt{e}$

2. (2) $\frac{3}{2} \sqrt{e}$

3. (3) $\frac{1}{2}+\sqrt{e}$

4. (4) $\frac{\sqrt{e}}{2}$

Correct Option: , 2

Solution:

$x^{3} d y+x y d y x=2 y d x+x^{2} d y$

$\Rightarrow\left(x^{3}-x^{2}\right) d y=(2-x) y d x$

$\Rightarrow \frac{d y}{y}=\frac{2-x}{x^{2}(x-1)} d x$

$\Rightarrow \int \frac{d y}{y}=\int \frac{2-x}{x^{2}(x-1)} d x$.....(i)

Let $\frac{2-x}{x^{2}(x-1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}$

$\Rightarrow 2-x=A(x-1)+B(x-1)+C x^{2}$

Compare the coefficients of $x, x^{2}$ and constant term.

$C=1, B=-2$ and $A=-1$

$\therefore \int \frac{d y}{y}=\int\left\{\frac{-1}{x}-\frac{2}{x^{2}}+\frac{1}{x-1}\right\} d x$

$\Rightarrow \ln y=-\ln x+\frac{2}{x}+\ln |x-1|+C$

$\therefore y(2)=e$

$\Rightarrow 1=-\ln 2+1+0+C \quad[\because \log e=1]$

$\Rightarrow C=\ln 2$

$\Rightarrow \ln y=-\ln x+\frac{2}{x}+\ln |x-1|+\ln 2$

At $x=4$

$\Rightarrow \ln y(4)=-\ln 4+\frac{1}{2}+\ln 3+\ln 2$

$\Rightarrow \ln y(4)=\ln \left(\frac{3}{2}\right)+\frac{1}{2}=\ln \left(\frac{3}{2} e^{1 / 2}\right)$

$[\because \log m+\log n=\log (m n)]$

$\Rightarrow y(4)=\frac{3}{2} e^{1 / 2}$